1. ## Integer numbers

Hi everyone.

I can't show this exercise:

Let $\displaystyle a,b\in{\mathbb{Z}$ be and $\displaystyle a<b$ then $\displaystyle a+1\leq{b}$

I've tried by contraction, but I haven't gotten anything.

Thanks

2. ## Re: Integer numbers

Well, I've done following:

Suppose that $\displaystyle a+1>b$, but $\displaystyle a<b\Leftrightarrow{a+1<b+1}$ hence it's a contraction because $\displaystyle a+1>b$ and $\displaystyle a+1<b+1$.

Therefore $\displaystyle a+1\leq{b}$

Is it correct?

3. ## Re: Integer numbers

Originally Posted by Fernando
Hi everyone.

I can't show this exercise:

Let $\displaystyle a,b\in{\mathbb{Z}$ be and $\displaystyle a<b$ then $\displaystyle a+1\leq{b}$

I've tried by contraction, but I haven't gotten anything.

Thanks
Suppose :

$\displaystyle a+1 > b$

hence :

$\displaystyle a>b-1 \Rightarrow a \geq b$

so we have contradiction because :

$\displaystyle a < b$

4. ## Re: Integer numbers

Originally Posted by princeps
Suppose :

$\displaystyle a+1 > b$

hence :

$\displaystyle a>b-1 \Rightarrow a \geq b$

so we have contradiction because :

$\displaystyle a < b$
Thanks! But is my answer correcto ?

5. ## Re: Integer numbers

Originally Posted by Fernando
Thanks! But is my answer correcto ?
Since :

$\displaystyle a+1 > b ~\text{and}~ a+1 < b+1$

cannot be true at the same time your reasoning is correct .

6. ## Re: Integer numbers

OK! thanks again !

7. ## Re: Integer numbers

Originally Posted by Fernando
Well, I've done following:
Suppose that $\displaystyle a+1>b$, but $\displaystyle a<b\Leftrightarrow{a+1<b+1}$ hence it's a contraction because $\displaystyle a+1>b$ and $\displaystyle a+1<b+1$.
Therefore $\displaystyle a+1\leq{b}$
Is it correct?
It is difficult to answer that. We don't even know what set of axioms are in use here.
Here is a standard theorem in the development of number systems.
Given any two numbers, if $\displaystyle b-a>1$ then $\displaystyle (\exists j\in\mathbb{Z})[a<j<b].$
That depends upon Archimedes' principle and well ordering.